Question

A tennis ball is released from rest at a height h = 32 m above the ground.
The ball collides with the ground and bounces up at 75 % of the impact
speed the ball had with the ground. The collision with the ground is
nearly instantaneously. A second ball is released above the first ball from
the same height the instant the first ball loses contact with the ground.
Calculate the time when the two balls collide in seconds.
(Take g = 9.8 m/s
2
)

Answer 1

Answer:

Let the two balls collide t sec after the first ball is thrown, and let h be the height at which they collide.

For the first ball:

h=40t−

2

1

gt

2

And for the second ball:

h=40(t−2)−

2

1

g(t−2)

2

40t−

2

1

gt

2

=40(t−2)−

2

1

g(t−2)

2

2gt=80+20=100

t=5 sec

∴h=40×5−

2

1

×10×5

2

=75m